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Mass spectrometer - no deflection

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    In one type of mass spectrometer the ions are passed through a velocity selector. This is a pair of parallel plates 2.5 cm apart. A pd of 140 V is applied to create an electric field. A magnetic field is also applied so as to give a force in the direction opposite to the field.
    Calculate the magnetic field strenght needed to allow uni-positive (loss of 1 electron) neon-20 ions (mass = 3.32 * 10-26) to pass un-deflected through a velcoity selector
    The neon-20 ions are moving at 3 * 105 ms-1

    2. Relevant equations

    3. The attempt at a solution
    So I think the way to do this is to equate
    BqV = mv2/r

    so B = mv2/rqV
    B = 10671.428... T
    (using a radius of 2.5/200)

    The answer given is 1.8 * 10-2 T
    How??
     
  2. jcsd
  3. Mar 27, 2012 #2
    In a velocity selector there are 2 field.... a magnetic field and an electric field between the parallel plates.
    The moving ion experiences a force due to the magnetic field and a force due to the electric field.
    What can you say about these forces if the ions are undeflected?
    Do you know how to calculate both forces?
     
  4. Mar 27, 2012 #3
    as the forces are ppd, I would say they would be deflected
    the only way there would be no deflection is if they acted along the same line but in opposite directions.

    anyway
    magnetic force = BqV
    E = Force/Q

    I presume your saying they are equal but surely it would still be deflected :S
     
  5. Mar 27, 2012 #4
    The fields are at right angles to each other....they are called 'crossed fields'
    You are correct to see that if there is no deflection then the 2 forces must be equal so....
    Bqv = Eq .....this means that v = E/B
    Can you find E and knowing v.....get the answer
     
    Last edited: Mar 27, 2012
  6. Mar 27, 2012 #5
    yes
    E = V/d
    Eq where q = 1.6 * 10-19
    q and V are known so we can find B

    but how to cross fields produce no resultant force?
     
  7. Mar 27, 2012 #6
    E = V/d = 140/0.025
    The force due to the electric field is in the opposite direction to the force due to the magnetic field. When these forces are equal there is no overall force and the ion beam is undeflected.
    you need to remembeer that the magnetic force is at right angles to the velocity of the ions
     
  8. Mar 27, 2012 #7
    ok so what would a diagram of velocity, electric force and magnetic force look like
    this would prob clear it up for me :)
     
  9. Mar 27, 2012 #8
    first of all.... did you get B = 0.018T?
    These 3 quantities are vectors at right angles to each other so if you draw v along the x axis and E along the y axis then B is along the z axis
     
  10. Mar 27, 2012 #9
    well I got 0.018666667 as the speed I gave was actually an approx (I worked out the actual speed in a previous part of the question)

    but if B and E are ppd to each other then the forces won't cancel, will they?

    so the electric force is in the same plane and same direction as E (up)

    as neon are +ive, we have (by flemming left)
    second finger to the right
    mag field into page and force upward
    so we have two upward forces?
     
  11. Mar 27, 2012 #10
    I got 0.0187T......that is the answer...just rounding off...not very important.
    Using your notation...+ ions moving to right, so second finger to right.
    B into paper so force due to magnetic field is up.
    This means that the + plate must be at the top and the - plate at the bottom to give the electric force down
     
  12. Mar 27, 2012 #11
    oh because it is not deflected, by definition the +ive plate MUST be at the top as the particle, due to e-field would move down?
     
  13. Mar 27, 2012 #12
    That's it...you have got it
     
  14. Mar 27, 2012 #13
    thanks, although this is what the textbook ms says:

    magnetic field perpendicularly out of page if top plate is +ive
    to check, does this assume particles are moving to the RIGHT?
     
  15. Mar 27, 2012 #14
    is there a diagram in the text book?
    If the field is out of the page and the +ions are moving to the LEFT then the + plate is at the top and the - plate at the bottom.
     
  16. Mar 27, 2012 #15
    thanks
    no diagram but I just wanted to check if I go it right :)
     
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